Integration

The Constant of Integration

When you integrate, you must always add + c at the end. This is called the constant of integration. We need it because when you differentiate any constant, you get zero, so we can't know what constant was originally there.

Example: If we differentiate x² + 5, we get 2x. If we differentiate x² + 100, we still get 2x. So when we integrate 2x, we write x² + c because c could be any number.

Note: You don't need + c when finding a definite integral (one with limits/numbers at the top and bottom of the integral sign).

1. Integrating e^(ax+b)

The exponential function e^x is special because when you differentiate it, you get itself back. For integration:

• ∫ e^x dx = e^x + c

When you have e^(ax+b) where a and b are constants:

• ∫ e^(ax+b) dx = (1/a) e^(ax+b) + c

The fraction 1/a appears because of the chain rule working in reverse. You divide by the coefficient of x.

Example: ∫ e^(3x+2) dx = (1/3) e^(3x+2) + c

Example: ∫ 2e^(5x) dx = 2 × (1/5) e^(5x) + c = (2/5) e^(5x) + c

2. Integrating 1/(ax+b)

The function 1/x integrates to give a logarithm (ln):

• ∫ (1/x) dx = ln|x| + c

The vertical bars | | mean "modulus" or "absolute value" — this just means we take the positive version of x. We need this because you can't take the logarithm of a negative number.

When you have 1/(ax+b):

• ∫ 1/(ax+b) dx = (1/a) ln|ax+b| + c

Again, divide by the coefficient of x.

Example: ∫ 1/(3x+2) dx = (1/3) ln|3x+2| + c

Example: ∫ 5/(2x-1) dx = 5 × (1/2) ln|2x-1| + c = (5/2) ln|2x-1| + c

Important: You cannot use the power rule (increasing the power by 1 and dividing) when the power is -1. That's why 1/x = x^(-1) needs this special logarithm rule.

3. Integrating sin(ax+b) and cos(ax+b)

For sine and cosine:

• ∫ sin x dx = -cos x + c (note the minus sign!)
• ∫ cos x dx = sin x + c

When you have sin(ax+b) or cos(ax+b):

• ∫ sin(ax+b) dx = -(1/a) cos(ax+b) + c
• ∫ cos(ax+b) dx = (1/a) sin(ax+b) + c

Example: ∫ sin(2x) dx = -(1/2) cos(2x) + c

Example: ∫ 3cos(4x+1) dx = 3 × (1/4) sin(4x+1) + c = (3/4) sin(4x+1) + c

Watch out: Don't forget the minus sign when integrating sin x!

5. Integrating 1/(x²+a²)

This one gives you an inverse tan function:

• ∫ 1/(1+x²) dx = tan^(-1) x + c
• ∫ 1/(a²+x²) dx = (1/a) tan^(-1)(x/a) + c

Example: ∫ 1/(4+x²) dx = ∫ 1/(2²+x²) dx = (1/2) tan^(-1)(x/2) + c

Example: ∫ 1/(2+3x²) dx

First, factor out the 3: = ∫ 1/(3(2/3 + x²)) dx = (1/3) ∫ 1/(2/3 + x²) dx

Now write 2/3 as (√(2/3))²: = (1/3) × (1/√(2/3)) tan^(-1)(x/√(2/3)) + c

This can be simplified to: (1/√6) tan^(-1)(x√(3/2)) + c

The "Adjust and Compensate" Technique

Sometimes you need to multiply and divide by the same number to get the integral in the right form. This is called "adjusting and compensating."

Example: ∫ 6e^(3x) dx

We know ∫ e^(3x) dx = (1/3)e^(3x) + c, but we have 6 in front.

Method: 6 = 2 × 3, so we can write: ∫ 6e^(3x) dx = ∫ 2 × 3e^(3x) dx = 2 ∫ 3e^(3x) dx = 2 × (1/3) × 3e^(3x) + c = 2e^(3x) + c

Or more directly: The coefficient of e^(3x) is 6, and we divide by the coefficient of x (which is 3): ∫ 6e^(3x) dx = (6/3)e^(3x) + c = 2e^(3x) + c